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8. An object is thrown up from a height of 5 meters with an initial speed of 10 m/s. By physical principles it is found that its position at any instant of time is given by: Y(t) = 5 + 10t − 4, 9t - And its speed at any instant of time has the form: V(t) = 10 - 9.8t. Calculate: A) the time of rise, t¸, knowing that at that instant of time the velocity of the object is zero, that is, v (ts) = 0. = B) the maximum height, that is, and (ts). C) return time, tv, knowing that the position coordinate at that instant is y (tv) = 0. = D) the speed with which it reaches the ground, i.e., v (tv).

8. An object is thrown up from a height of 5 meters with an initial speed of 10 m/s. By physical principles it is found that its position at any instant of time is given by: Y(t) = 5 + 10t − 4, 9t - And its speed at any instant of time has the form: V(t) = 10 - 9.8t. Calculate: A) the time of rise, t¸, knowing that at that instant of time the velocity of the object is zero, that is, v (ts) = 0. = B) the maximum height, that is, and (ts). C) return time, tv, knowing that the position coordinate at that instant is y (tv) = 0. = D) the speed with which it reaches the ground, i.e., v (tv).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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8. An object is thrown up from a height of 5 meters with an initial speed of 10 m/s. By physical principles it is found that its position at any instant of time is given by: Y(t) = 5 + 10t − 4, 9t - And its speed at any instant of time has the form: V(t) = 10 - 9.8t. Calculate: A) the time of rise, t¸, knowing that at that instant of time the velocity of the object is zero, that is, v (ts) = 0. = B) the maximum height, that is, and (ts). C) return time, tv, knowing that the position coordinate at that instant is y (tv) = 0. = D) the speed with which it reaches the ground, i.e., v (tv).
Transcribed Image Text:8. An object is thrown up from a height of 5 meters with an initial speed of 10 m/s. By physical principles it is found that its position at any instant of time is given by: Y(t) = 5 + 10t − 4, 9t - And its speed at any instant of time has the form: V(t) = 10 - 9.8t. Calculate: A) the time of rise, t¸, knowing that at that instant of time the velocity of the object is zero, that is, v (ts) = 0. = B) the maximum height, that is, and (ts). C) return time, tv, knowing that the position coordinate at that instant is y (tv) = 0. = D) the speed with which it reaches the ground, i.e., v (tv).
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